$\newcommand{\bf}[1]{\mathbb #1}\newcommand{\sc}[1]{\mathscr #1}$
A *duality* between two vector spaces $E$ and $F$ over $\bf K$ ($= {\bf R}$ of ${\bf C}$)
is, by definition, a bilinear form
$$
\langle \cdot, \cdot\rangle :E\times F\to \bf K,
$$
such that, if $\langle x, y\rangle =0$ for every $x$ in $E$, then $y=0$. And vice-versa.

Given a duality as above, one defines the *weak
topology* on $F$, usually denoted $\sigma (F,E)$, to be the coarsest topology according to which the linear functionals
$$
y\in F\mapsto \langle x, y\rangle \in \bf K
$$
are continuous for every $x$ in $E$.

It is a classical fact that every $\sigma (F,E)$-continuous linear functional $\varphi :F\to \bf K$, may be *represented by* a vector in
$E$ in the sense that there exists a (necessarily unique) $x$ in
$E$ such that
$$
\phi(y) = \langle x, y\rangle ,\quad\forall y\in E.
$$

One could therefore ask:

**Question**. Does the above still hold if continuity is replaced by *sequential continuity*. In other words,
must every sequentially $\sigma (F, E)$-continuous linear functional on $F$ be represented by a vector in $E$.

Before the reader jumps to the task of proving or disproving it, let me say that unfortunately the answer is negative, a counter example being presented below.

So let me specialize this a bit by restricting to the situation in which $E$ is a Banach space and $F$ is its topological dual, with the canonical duality $$ \langle x, \varphi \rangle = \varphi (x), \quad \forall x\in E, \quad \forall \varphi \in E'. $$

To be precise:

**Question**. Let $E$ be a Banach space and let $\varphi $ be a linear functional on $E'$ which is sequentially
$\sigma (E',E)$-continuous. Is $\varphi $ necessarily represented by a vector in $E$?

This is obviously true if $E$ is reflexive and I think I can also prove it for $E=c_0$, as well as for $E=\ell ^1$.

**A COUNTER EXAMPLE**

Let $E=\sc F(H)$ be the set of all finite-rank operators on Hilbert's space, and $F=\sc B(H)$, with duality defined by means of the trace, namely $$ \langle S, T\rangle = \text{tr}(ST), \quad\forall S\in \sc F(H), \quad\forall T\in \sc B(H). $$

In this case $\sigma \big (\sc B(H),\sc F(H)\big )$ turns out to be the weak operator topology (WOT), which coincides with the sigma weak operator topology ($\sigma $-WOT) on bounded subsets of $\sc B(H)$.

Since WOT-convergent sequences are bounded by Banach-Steinhauss, we have that the WOT-convergent sequences are the same as the $\sigma $-WOT convergent ones. It follows that every $\sigma $-WOT-continuous linear functional on $\sc B(H)$ is also WOT-continuous. Making a long story short, for every trace class operator $S$ on $H$ of infinite rank, the linear functional $$ T\in \sc B(H) \mapsto \text{tr}(ST)\in {\bf C} $$ is sequentially WOT-continuous, but it is not represented by an operator in $\sc F(H)$.

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